How do you find the axis of symmetry and vertex point of the function: y=x^2+6x-2?

Oct 9, 2015

The method is called completing the square

Explanation:

We have ${x}^{2} + 6 x - 2$
in the form $a {x}^{2} + b x + c$
where $a = 1$, $b = 6$, and $c = - 2$

So we divide the $b$ value by $2$ to get $3$, and square it but keep it as ${3}^{2}$, and add it to the $b$ value and minus it from the $c$ value.

${x}^{2} + 6 x + {3}^{2} - 2 - {3}^{2}$

so ${x}^{2} + 6 x + {3}^{2}$ is basically ${\left(x + 3\right)}^{2}$. So shorten it that way and we have..

${\left(x + 3\right)}^{2}$ and $- 2 - \left(9\right)$
So that is, ${\left(x + 3\right)}^{2} - 11$
From this, $x + 3 = 0$ and $x = - 3$. and $y = - 11$.
Line of symmetry is equal to $x = - 3$.

Oct 9, 2015

Find vertex of $y = {x}^{2} + 6 x - 2$

Ans: (-3, -11)

Explanation:

$y = {x}^{2} + 6 x - 2.$
x-coordinate of vertex:
$x = \left(- \frac{b}{2 a}\right) = \frac{- 6}{2} = - 3$
y-coordinate of vertex:
y = y(-3) = 9 - 18 - 2 = - 11
Vertex (-3, -11)
graph{x^2 + 6x - 2 [-40, 40, -20, 20]}